//等于i的区间
{x | i(i-1)/2 + 1 <= x <= i(i+1)/2}

//推出
{i | sqrt(2x+0.25)-0.5 <= i <= sqrt(2x-1.25)+0.5 }

//验算下左右的值，发现取右，所以
i = floor(sqrt(2x-1.25)+0.5);